hdu4292（最大流）

Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4415    Accepted Submission(s): 1492

Problem Description

　　You, a part-time dining service worker in your college’s dining hall, are now confused with a new problem: serve as many people as possible.
　　The issue comes up as people in your college are more and more difficult to serve with meal: They eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly.
　　You have prepared F (1 <= F <= 200) kinds of food and D (1 <= D <= 200) kinds of drink. Each kind of food or drink has certain amount, that is, how many people could this food or drink serve. Besides, You know there’re N (1 <= N <= 200) people and you too can tell people’s personal preference for food and drink.
　　Back to your goal: to serve as many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service.

 

Input

　　There are several test cases.
　　For each test case, the first line contains three numbers: N,F,D, denoting the number of people, food, and drink.
　　The second line contains F integers, the ith number of which denotes amount of representative food.
　　The third line contains D integers, the ith number of which denotes amount of representative drink.
　　Following is N line, each consisting of a string of length F. e jth character in the ith one of these lines denotes whether people i would accept food j. “Y” for yes and “N” for no.
　　Following is N line, each consisting of a string of length D. e jth character in the ith one of these lines denotes whether people i would accept drink j. “Y” for yes and “N” for no.
　　Please process until EOF (End Of File).

 

Output

　　For each test case, please print a single line with one integer, the maximum number of people to be satisfied.

 

Sample Input

   
   
   
   

    
    
    
    4 3 3
1 1 1
1 1 1
YYN
NYY
YNY
YNY
YNY
YYN
YYN
NNY
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3
   
   
   
   

题意：

    有F种食物和D种饮料，每种食物或饮料只能供有限次，且每个人只享用一种食物和一种饮料。现在有n头牛，每个人都有自己喜欢的食物种类列表和饮料种类列表，问最多能使几个人同时享用到自己喜欢的食物和饮料。

思路：

由于又有食物又有饮料，将每个人人拆分成2个，然后食物连第一部分人，第二部分人连饮料，第一部分人对应连上第二部分人，然后食物和饮料的数量放在与超级源点连接的边上。

总结：

tle的原因又莫名其妙的多了一种，边的最大值mm一定要足够大！！！刚开始没改边的最大值导致tle了，搞得莫名其妙的。。。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
typedef long long ll;
const   int oo=1e9;
/**oo 表示无穷大*/
const  int mm=111111111;
/**mm 表示边的最大数量，记住要是原图的两倍，在加边的时候都是双向的*/
const  int mn=999;
/**mn 表示点的最大数量*/
int node,src,dest,edge;
/**node 表示节点数，src 表示源点，dest 表示汇点，edge 统计边数*/
int ver[mm],flow[mm],nex[mm];
/**ver 边指向的节点，flow 边的容量 ，next 链表的下一条边*/
int head[mn],work[mn],dis[mn],q[mn];
void prepare(int _node, int _src,int _dest)
{
    node=_node,src=_src,dest=_dest;
    for(int i=0; i<node; ++i)head[i]=-1;
    edge=0;
}
/**增加一条 u 到 v 容量为 c 的边*/
void addedge( int u,  int v,  int c)
{
    ver[edge]=v,flow[edge]=c,nex[edge]=head[u],head[u]=edge++;
    ver[edge]=u,flow[edge]=0,nex[edge]=head[v],head[v]=edge++;
}
/**广搜计算出每个点与源点的最短距离，如果不能到达汇点说明算法结束*/
bool Dinic_bfs()
{
    int i,u,v,l,r=0;
    for(i=0; i<node; ++i)dis[i]=-1;
    dis[q[r++]=src]=0;
    for(l=0; l<r; ++l)
        for(i=head[u=q[l]]; i>=0; i=nex[i])
            if(flow[i]&&dis[v=ver[i]]<0)
            {
                /**这条边必须有剩余容量*/
                dis[q[r++]=v]=dis[u]+1;
                if(v==dest)  return 1;
            }
    return 0;
}
/**寻找可行流的增广路算法，按节点的距离来找，加快速度*/
int Dinic_dfs(  int u, int exp)
{
    if(u==dest)  return exp;
    /**work 是临时链表头，这里用 i 引用它，这样寻找过的边不再寻找*/
    for(  int &i=work[u],v,tmp; i>=0; i=nex[i])
        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
        {
            flow[i]-=tmp;
            flow[i^1]+=tmp;
            /**正反向边容量改变*/
            return tmp;
        }
    return 0;
}

int Dinic_flow()
{
    int i,ret=0,delta;
    while(Dinic_bfs())
    {
        for(i=0; i<node; ++i)work[i]=head[i];
        while((delta=Dinic_dfs(src,oo)))ret+=delta;
    }
    return ret;
}
char ss[1005];
int main()
{
    int n,d,f;
    while(~scanf("%d%d%d",&n,&f,&d))
    {
        int s=0,t=n*2+f+d+1,x;
        prepare(t+1,s,t);
        for(int i=1; i<=f; i++)
        {
            scanf("%d",&x);
            addedge(0,i,x);
        }
        int bj=f+2*n;
        for(int i=1; i<=d; i++)
        {
            scanf("%d",&x);
            addedge(bj+i,t,x);
        }
        for(int i=1; i<=n; i++)
        {
            addedge(f+i,f+n+i,1);
            scanf("%s",ss);
            for(int j=1; j<=f; j++)
                if(ss[j-1]=='Y')
                    addedge(j,f+i,1);
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%s",ss);
            for(int j=1; j<=d; j++)
                if(ss[j-1]=='Y')
                    addedge(f+n+i,bj+j,1);
        }
        printf("%d\n",Dinic_flow());
    }
}